c字串:如果在句子中找到一个词。-cstring:put''ifawordfoundinthesentence

0  

In C strings are terminated with a null character '\0'. Your code defines a somehow random number at the beginning (B and S) and iterates over that much characters instead of the exact number of characters, the strings actually contain. You can use the fact that the string is terminated by testing the content of the string in a while loop.

在C字符串中，以null字符“\0”结束。您的代码在开头(B和S)定义了一个任意的随机数，并遍历了许多字符，而不是字符的确切数目，字符串实际上包含了。您可以使用一个事实，即字符串通过在while循环中测试字符串的内容而终止。

i = 0;
while (str[i]) {
  ...
  i = i + 1;
}

If you prefer for loops you can write it also as a for loop.

如果你喜欢循环，你也可以把它写成for循环。

for (i = 0; str[i]; i++) {
  ...
}

Your code does not move the contents of the remaining string to the left. If you replace two characters ol with one character , you have to move the remaining characters to the left by one character. Otherwise you would have a hole in the string.

您的代码不会将剩余字符串的内容移动到左侧。如果你用一个字符替换两个字符，你必须将剩下的字符移到一个字符的左边。否则你会在弦上有个洞。

#include 

int main() {
  char small[] = "ol";
  char big[] = "my older gradmom see my older sister";
  int s; // index, which loops through the small string
  int b; // index, which loops through the big string
  int m; // index, which loops through the characters to be modified

  // The following loops through the big string up to the terminating
  // null character in the big string.
  b = 0;
  while (big[b]) {
    // The following loops through the small string up to the
    // terminating null character, if the character in the small
    // string matches the corresponding character in the big string.
    s = 0;
    while (small[s] && big[b+s] == small[s]) {
      // In case of a match, continue with the next character in the
      // small string.
      s = s + 1;
    }
    // If we are at the end of the small string, we found in the
    // big string.
    if (small[s] == '\0') {
      // Now we have to modify the big string.  The modification
      // starts at the current position in the big string.
      m = b;
      // First we have to put the space at the current position in the
      // big string.
      big[m] = ' ';
      // And next the rest of the big string has to be moved left. The
      // rest of the big string starts, where the match has ended.
      while (big[b+s]) {
        m = m + 1;
        big[m] = big[b+s];
        s = s + 1;
      }
      // Finally the big string has to be terminated by a null
      // character.
      big[m+1] = '\0';
    }
    // Continue at next character in big string.
    b = b + 1;
  }
  puts(big);
  return 0;
}